$\min$ of expression $\sin \alpha+\sin \beta+\sin \gamma,$satisfying $\alpha+\beta+\gamma = \pi$ 
The $\min$ of expression $\sin \alpha+\sin \beta+\sin \gamma,$ Where $\alpha,\beta,\gamma\in \mathbb{R}$ satisfying $\alpha+\beta+\gamma = \pi$

$\bf{Options ::}$ $(a)\;\; + ve \;\;\;\;\;\;\; (b)\;\; -ve \;\;\;\;\;\; (c)\;\; 0\;\;\;\;\;\; (d)\;\; -3$
$\bf{My\; Try::}$ Putting $\displaystyle \alpha = -\frac{\pi}{2}\;\;\;\;\;\; ,\beta = -\frac{\pi}{2}\;\;\;\;\;\; ,\gamma = 2\pi$
Then we get $\sin\alpha+\sin \beta+\sin \gamma = -ve $
$\bf{Added::}$ Trying using analytical way: Given $\alpha+\beta+\gamma = \pi$
$$\sin \alpha+\sin \beta+\sin \gamma = 2\sin \left(\frac{\alpha+\beta}{2}\right)\cos \left(\frac{\alpha-\beta}{2}\right)+2\sin \frac{\gamma}{2}\cdot \cos \frac{\gamma}{2}$$
So $$ = 2\cos \frac{\gamma}{2}\left[\cos \left(\frac{\alpha-\beta}{2}\right)+\cos \left(\frac{\alpha+\beta}{2}\right)\right] = 4\cos \frac{\alpha}{2}\cos \frac{\beta}{2}\cos \frac{\gamma}{2}$$
Now Using $\displaystyle -1 \leq \cos \frac{\alpha}{2}\;,\cos \frac{\beta}{2}\;,\cos \frac{\gamma}{2}\leq 1$
So we get $$\sin \alpha+\sin \beta+\sin \gamma = 4\cos \frac{\alpha}{2}\cos \frac{\beta}{2}\cos \frac{\gamma}{2}\geq -4$$
But this is not possible bcz $\displaystyle \cos \frac{\alpha}{2}\;,\cos \frac{\beta}{2}\;,\cos \frac{\gamma}{2}\neq -1$ Simultaneously 
I did not understand how can i solve it
Help me , Thanks
 A: 

There is a nice geometric way for this problem. Note: It comes from one of the IMO problem and is not my original idea.
(This paragraph corresponds to the first picture)
In triangle $ABC$, the three altitudes $AA'$, $BB'$ and $CC'$ intersect at orthocenter $H$. Denote the midpoint of $AH, BH, CH$ as $A'', B'', C''$. It is fairly obvious that area of the shape $AC'A''B'$ is same as area of the shape $HC'A''B'$. Now by symmetry over the other two vertices as well and sum them up, we can conclude the hexagon $A''B'C''A'B''C'$ has exactly half the area of triangle $ABC$. Furthermore, we can see $B'A'' = C'A'' = {1\over 2}AH$.
(This paragraph corresponds to the second picture)
Now draw the three altitudes inside triangle $A'B'C'$, after some angle tracing we can find out $\angle B'A''C'=\angle B'MC'$ so if you imagine mirroring point A'' along axis $B'C'$ then the two angles would be on the same circle, with $M$ closer to (at most the same distance) $B'C'$ than the mirror of $A''$ because $A''B'=A''C'$. So the area of triangle $B'A''C'$ is larger than or equal to the area of $B'MC'$. Apply this over the other two vertices and sum them up we can conclude area of $A'B'C'$ is at most $1\over 2$ of the hexagon and is at most $1\over 4$ of the original triangle $ABC$.
Now this means the sum of areas of $AB'C', BC'A', CA'B'$ is at least $3\over 4$ of triangle $ABC$. All three triangles are similar to $ABC$ and the side ratios are $cosA, cosB, cosC$ respectively. Therefore we can conclude
$$cos^2A+cos^2B+cos^2C \geq {3\over 4}$$
Using the identity $cos^2x+sin^2x=1$, we get
$$sin^2A+sin^2B+sin^2C \leq {9\over 4}$$
Therefore,
$$(sinA+sinB+sinC)^2\leq 3(sin^2A+sin^2B+sin^2C) \leq {27\over 4}$$
By Cauchy Inequality. Therefore
$$-{3\sqrt{3} \over 2}\leq sinA+sinB+sinC \leq {3\sqrt{3} \over 2}$$
Note that negative angles can also be represented by the original triangle if we consider the three sides $AB, BC, CA$ as vectors and flipping positions of $A,B,C$. The key point is the area ratio of $AB'C'$ to $ABC$ is always $cos^2A$
A: As there was no solution to this old problem (more than two years old), I have written this one. I find back a solution mentionned by @Michael Hoppe.
I transform expression $\sin(\alpha)+\sin(\beta)+\sin(\gamma)$ with constraint $\alpha+\beta+\gamma=1$ under the form :
$$\sin(\alpha)+\sin(\beta)+\sin(\pi -(\alpha+\beta))$$
i.e.,
$$f(\alpha,\beta):=\sin(\alpha)+\sin(\beta)+\sin(\alpha+\beta).$$
A basic patch of this doubly periodical function (separately periodical for variable $\alpha$ and variable $\beta$) is represented on Fig. 1. In fact the minimum must occur along the intersection of the surface with vertical plane with equation $y=x$ (see Appendix below).
Thus it suffices to study the minimum of function 
$$\varphi(\theta):=f(\theta,\theta)=2 \sin(\theta)+\sin(2 \theta).$$
Let us determine the extrema of $\varphi$. The occur for values of $\alpha$ such that :
$$\varphi'(\theta)=2 \cos(\theta)+2 \cos(2 \theta)=0$$
This equation can be transformed into $\cos(\theta)=-\cos(2 \theta)$ then into $\cos(\theta)=\cos(\pi-2 \theta)$ whence 2 cases :


*

*(a) $\theta=\pi-2\theta + k 2 \pi$ yielding $\theta=\pi/3+k 2 \pi/3$, or

*(b) $\theta=-(\pi-2\theta) + k 2 \pi$ yielding $\theta=\pi + k 2 \pi$ (that is in fact included into case (a)).
It means that there are potentially 3 solutions :
$$\theta_1=\pi/3, \ \ \ \theta_2=\pi, \ \ \ \theta_3=5 \pi/3.$$
In fact, as can be verified on the figure, case  $\theta_1$ corresponds to a maximum and case  $\theta_2$ corresponds to an inflexion point (a stationnary point in 3D). Only case $\theta_3=5 \pi/3$ gives a minimum.
Thus, the solution is for any $k \in \mathbb{Z}$
$$\alpha=\beta=5\pi/3+k 2 \pi, \gamma=\pi-(\alpha+\beta)=-7 \pi/3 - k 2 \pi$$
that can be given the following simpler equivalent form :

$$\alpha=\beta=-\pi/3+k 2 \pi, \gamma=\pi-(\alpha+\beta)=5\pi/3 - k 4 \pi$$


Fig. 1 : Surface $z=f(x,y)$ with curve of function $\varphi$ traced on it.

Appendix : Why do the minimum must appear for values of $\alpha, \beta$ such that $\alpha=\beta$ ? 
Let us work with $\alpha+\beta=const. = c$ (up to a multiple of $2 \pi$).
If such is the case, differentiating $f(\alpha,\beta)=\sin(\alpha)+\sin(c-\alpha)+\sin(c)$ gives $\cos(\alpha)-\cos(c-\alpha)$ which is $0$ iff 


*

*either $\alpha=c - \alpha +k 2 \pi \ \iff \ \alpha=\frac{c}{2}+k \pi$, meaning that $\alpha=\beta=\frac{c}{2}+k \pi$.

*or $\alpha=-(c - \alpha) +k 2 \pi$ which happens in the very particular case where $c=k 2 \pi$.


Edit : it is possible that there is a solution using the (classical) identity you mention
$$f(\alpha,\beta,\gamma):=\sin \alpha+\sin \beta+\sin \gamma = 4\cos \frac{\alpha}{2}\cos \frac{\beta}{2}\cos \frac{\gamma}{2}$$
valid under the constraint $\alpha+\beta+\gamma=\pi$... but I don't see how...
A: With an hopefully obvious notation, $$\sin\alpha+\sin\beta+\sin(\alpha+\beta)=a+b\pm a\sqrt{1-b^2}\pm b\sqrt{1-a^2}.$$
Cancelling the gradient, and rearranging,
$$\begin{cases}\sqrt{1-a^2}\pm\sqrt{1-a^2}\sqrt{1-b^2}\mp ab=0,
\\\sqrt{1-b^2}\mp ab\pm\sqrt{1-a^2}\sqrt{1-b^2}=0.\end{cases}$$
This proves a symmetry, $\sqrt{1-a^2}=\sqrt{1-b^2}$ or $a=\pm b$.
Now,
$$\sqrt{1-a^2}\pm(1-a^2)\mp a^2=0$$ or 
$$c\pm c^2\mp(1-c^2)=0$$
yields the positive solutions
$$c=1,a=0
\\c=\frac12,a=\pm\frac{\sqrt3}2.$$
and the minimum is
$$-\frac{3\sqrt3}2.$$
